In the given circuit- a charge of -80mu C is given to the upper plate of the 4 mu F capacitor- Then in the steady state- the charge on the upper plate of the 3 mu F capacitor is-32mu C-40mu C-48mu C-80mu C

Question

Toppr Admin · Accepted Answer

Let q be the charge on 3-x3BC-F capacitor-xA0-then-x27F9-80-x2212-q2-q3q-48-x3BC-C

In the given circuit, a charge of $+ 80 μ C$ is given to the upper plate of the $4 μ F$ capacitor. Then in the steady state, the charge on the upper plate of the $3 μ F$ capacitor is

$+ 32 μ C$
$+ 40 μ C$
$+ 48 μ C$
$+ 80 μ C$

let q be the charge on $3 μ F$ capacitor then
$⟹ \frac{80 - q}{2} = \frac{q}{3}$
$q = 48 μ C$

In the given circuit, a charge of +80μC is given to the upper plate of the 4 μF capacitor. Then in the steady state, the charge on the upper plate of the 3 μF capacitor is+32μC+40μC+48μC+80μC

let q be the charge on 3μF capacitor then⟹80−q2=q3q=48μC

In the given circuit, a charge of $+ 80 μ C$ is given to the upper plate of the $4 μ F$ capacitor. Then in the steady state, the charge on the upper plate of the $3 μ F$ capacitor is

$+ 32 μ C$
$+ 40 μ C$
$+ 48 μ C$
$+ 80 μ C$

let q be the charge on $3 μ F$ capacitor then
$⟹ \frac{80 - q}{2} = \frac{q}{3}$
$q = 48 μ C$